-16t^2+48t+2-38=0

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Solution for -16t^2+48t+2-38=0 equation: 



-16t^2+48t+2-38=0

We add all the numbers together, and all the variables

-16t^2+48t-36=0

a = -16; b = 48; c = -36;
Δ = b2-4ac
Δ = 482-4·(-16)·(-36)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{-48}{-32}=1+1/2$

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